I won’t be in on Monday (Yom Kippur). Prof. Ravi
Montenegro will run the class that day.
If you have questions about past homework problems, I’m
happy to answer them in class, but it’s best if you email me
your question so that I can prepare an answer for class. (I
might also just answer by email.)
Questions on material up through section 2.2?
Main ideas of Section 2.3? …
Every convergent sequence is bounded.
Algebraic limit theorem (“limits respect +, –, , and ”)
Order limit theorem (“limits respect  and ”)
In high school geometry and algebra, when you wanted to
prove something of the form “P implies Q”, you got good
at constructing proofs of the type “If P is true, then P is
true, so P is true, so P is true, which implies that Q is
true, which is what we wanted to prove.” And if what you
were trying to prove was algebraic, you could usually leave
out all the words and just write down equations and
inequalities.
These proofs in this class aren’t like that, even when the
theorem is of the form “P implies Q”, because typically
both the premise P and the conclusion Q involve
quantifiers.
E.g., say we’re trying to prove “If (xn) is convergent then
(xn) is bounded” (Theorem 2.3.2). That is, we’re trying to
show that
(L)( > 0)(N)(n  N) |xn–L| < 
implies
(M)(n) |xn|  M.
If you remove all the words from Abbott’s proof, all you’re
left with is the inscrutable sequence of formulas
=1
NN
nN
xn  (L–1, L+1)
|xn| < |L| + 1
M = max{|x1|, |x2|, |x3|, …, |xN–1|, |L|+1}
|xn|  M
It’s the words that give these formulas their context and
turn them into a proof.
N.B.: There is a huge difference between “WE KNOW that
P is true, THEREFORE WE KNOW that Q is true” and
“WE WANT P to be true, AND WE CAN MAKE P TRUE
by making Q true”. Simply writing “P, so Q” could mean
either!
So, knowing how to connect the formulas with words is a
crucial part of proofcraft.
There can be more than one way to write a valid proof,
depending on how much of the thought processes that went
into the proof you want to display.
Here’s an example of this:
Theorem: If lim an = a, then lim 2an = 2a. (This is
Theorem 2.3.3(i) in the special case c=2.)
Proof #1: Given any  > 0, there exists N such that |an – a| <
/2 whenever n  N (because lim an = a), so |2an – 2a| < 
whenever n  N, which implies lim 2an = 2a (by the
definition of convergence). 
Proof #2 (in the style of Abbott, following the template he
described in Section 2.2). Let  be some arbitrary positive
number. Our goal is to find some point in the sequence
(2an) after which we have
|2an – 2a| < .
Now,
|2an – 2a| = |2(an – a)| = |2| |an – a| = 2 |an – a|.
We are given that (an)  a so we can make |an – a| as small
as we like. In particular we can choose N such that
|an – a| < /2
whenever n  N. To see that this N indeed works, observe
that, for all n  N,
|2an – 2a| = 2 |an – a| < 2 (/2) = . 
Both of these are fine.
Questions on pages 44 – 49?
Exercise 2.3.7(a): Prove that if (an) is bounded and (bn)
converges to 0, then (anbn) converges to 0.
Can we derive this from the Algebraic Limit Theorem? …
We could if we knew that (bn) converges to some limit L.
But we’re not told that (bn) converges.
Proof: Since (an) is bounded, there exists M such that |an| 
M.
For any  > 0, there exists N such that for all n  N, |bn – 0|
<  / M, implying that |anbn – 0| = |an| |bn| < (M) ( / M) = .
Since  was arbitrary, we have proved that lim anbn = 0. 
Exercise 2.3.7(c): Use part (a) to prove Theorem 2.3.3(iii)
in the case when a = 0. That is, show that if lim an = a = 0
and lim bn = b, then lim anbn = ab = 0.
Solution: Since (an) converges to 0 and (bn) is bounded
(why? …
by Theorem 2.3.2), (anbn) converges to 0.
Exercise 2.3.3 (Squeeze Theorem): Show that if xn  yn  zn
for all n  N, and if lim xn = lim zn = L, then lim yn = L.
Can we derive this from the Order Limit Theorem? …
We could if we knew that (yn) converges.
But we’re not told ahead of time that (yn) converges.
Proof: Given  > 0, …
[use the hypotheses] …
there exist N1 and N2 such that for all n  N1, |xn – L| < ,
and such that for all n  N2, |zn – L| < .
Let N = …
max(N1, N2).
Then for all n  N, L –  < xn < L +  and L –  < zn < L + ,
so L –  < xn  yn  zn < L + , so |yn – L| < .
Since  was arbitrary, we have proved that lim yn = L. 